ME 2304: 3D Geometry & Vector Calculus Dr. Faraz Junejo

ME 2304: 3D Geometry & Vector Calculus

Dr. Faraz Junejo

Multiplication of a Vector

By a scalar b = m a

a

Magnitude m a

Direction along a

b

We can increase or decrease the magnitude of a vector by multiplying the vector by a scalar

Scalar Multiplication

In the examples, vector B (2 units) is half the size of vector A (which is 4 units) . We can write:

B = 0.5 A This is an example of a scalar multiple. We have

multiplied the vector A by the scalar 0.5.

ExerciseGiven a = (15,-6,24) , b = (5,-2, 8) and c = (-15/2, 3, -12).a. Show that a & b have the same direction.b. Show that a & c have opposite direction

By inspecting a & b, it can be seen that: a = 3b or b = 1/3 a

Since the scalar 3 or 1/3 is positive, this implies a & b have the same direction.

Whereas, by inspecting a & c, it can be seen that: a = -2c or c = -1/2 a

Since the scalar -2 or -1/2 is negative, this implies a & c have opposite directions.

Multiplication of Vectors(contd)By a vector

two types

Dot (Scalar) Product

Cross (Vector) Product

Scalar Product

a

b

multiplication SCALAR

a . b

DOT PRODUCT

Dot Product (contd)

a . b

a

b

= a b cos 360 -

In words, a.b is the length of a times the length of b times the cosine of the angle between a and b

Dot Product (contd)Magnitude of either one ofthe vectors

Component of the other along the direction of the first

X

a

b

bcos a

b acos

Dot Product: SummaryIf we have any two vectors, say a and b, the dot product of a and b is given by:

cosbaba

Where, |a| and |b| are the magnitudes of a and b respectively, and θ is the angle between the 2 vectors.

The dot product of the vectors a and b is also known as the scalar product since it always returns a scalar value.

W = F . d

Work done on a particle by a

constant force F during a

displacement d is

Examples in Physics

Exercise

A small cart weighing 100 lb is pushed up an incline that makes an angle of 30 Degrees with the horizontal. Find the work done against gravity in pushing the cart a distance of 80ft.

80 ft

Θ=30

R (a1,a2)

T xP

Q (0,-100)

y

The vector PQ represents the force of gravity acting vertically downward with a magnitude of 100 lb.

The corresponding vector PQ is (0i-100j)

From triangle PTR, we conclude that 403080a , 34030cos80 21 Sina

jiThus 40340PR , Therefore, the work done against gravity is PQ.PR =

4000 ft-lb

Find the dot product of force vectors F1= 4N and F2= 6N acting at 40 Degrees to each other ?

F1=4N

F2=6N

θ

cos2121 FFFF

40)6)(4(21 CosFF

NFF 38.1821

Scalar Product: Applications

To test whether two vectors are perpendicular.

If the dot product of two non zero vectors is zero, the vectors are perpendicular.

0143

)k1j2i()k1j2i(3

ˆˆˆˆˆˆ

Finding the angle between two vectors.

ba

ba

cos

Scalar Product: Applications

cosbaba

ba

ba

1cos

kj2i2a ˆˆˆ

k2j3i6b ˆˆˆ

79)21

4(cos

7.3

4cos 1

What is the angle between a and b ?

39122 222 a

749236 222 b

42612 ba

Find the dot product of vectors P and Q, if they are acting at right angle to each other and

Exercises

,5,7 QP

Answer: 0

Find P.Q , if P= 6i+5j and Q= 2i-8jAnswer: -28

Find the angle between P=3i-5j and Q=4i+6j. Draw a rough sketch to verify your answer.

Answer: 115.3 Degrees

If, the work done by F in acting from P to q is

60 and ,3,40 mDNF

Answer: 60 Joules

b

Scalar Product: Applications

Q

P

XY

a

Finding the projection of one vector on another vector.

Projection of a on b is XY The projection of a onto b will be

given by:

b

b cosaaprojb

In summary, the proja b has length

cosa , and directionb

b

It is called the scalar component of a in the direction of b

Vector Projection (contd)

baba cos

Therefore, scalar component of a in the direction of b

b

baa

cos

kj2i2a ˆˆˆ

k2j3i6b ˆˆˆ

Vector Projection (contd)What is the scalar component of a in the direction of b ?

Scalar component of a on b is

7

4

b

ba

Since, projection of a onto b is:

b

b cosaaprojb

we know,

b

baa

cos

Therefore,

bb

ba

b

b

b

baaprojb

2

Similarly, projection of b onto a is given by:

aa

abbproja

2

b onto a of Projection

• If b is a vector not equal to zero, then any vector “a” can be projected onto “b” (as mentioned earlier) as well as onto a vector

• It can be seen from figure shown below that using vector addition it can be written:

b magnitude of b

a

b

b

aojbPr

aojb

Pr

aProjaaProj

aProjaProja

bb

bb

Find the vector projection of b = (6i+3j+2k) onto a=i-2j-2k and the scalar component of b in the direction of a

)22()2()2()1(

4662

2222 kjia

a

abbproja

kjikjibproja 9

8

9

8

9

4)22(

9

4

Now the scalar component of b in the direction of a will be given by:

3

4

3

466cos

a

abb

Cross Product

a

b a b

CROSS PRODUCT

multiplication VECTOR

Vector Product (contd)

a

b

Magnitude : Area of the parallelogram generated by a and b.

Vector Product (contd)

a

b

Magnitude : sinabah

sinbh

Vector Product (contd)

a

b

ba

sinbh

Direction : Perpendicular to both a and b.

Right-Hand Rule

The order of vector multiplicationis important.

Geometrical Interpretation

a

b

b sin

sin|ba| ab

A = a b

Area of the parallelogram formed by a being the base and height being b sin

Cross Product: SummarySuppose we have 2 vectors a and b. These 2 vectors lie on a plane and the unit vector n is normal (at right angles) to that plane

The cross product (also known as the vector product) of a and b is given by:

sinbaba

The right hand side represents a vector at right angles to the plane containing vectors A and B.

Cross Product: Properties

00sin|ba| ab

and a and b are not null vectors, then a is parallel to b.

If 0ba

Cross Product: Determinant

321

321

kji

bbb

aaaba

• The cross product can be expressed as

• Expanding the determinants gives

ˆ ˆ ˆ

ˆ ˆ ˆy z x yx zx y z

y z x yx zx y z

A A A AA AA A A

B B B BB BB B B

i j k

A B i j k

ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k

Vector Products Using Determinants

ExampleA simple cross product

132

215

ˆˆˆ

kji

k

j

i

ˆ)]1.2(3.5[

ˆ)]1.5(2.2[

ˆ]2.3)1.1[(

-5

)ˆˆ3ˆ2()ˆ2ˆˆ5( kjikji

9

17

kji ˆ17ˆ9ˆ5

CHECK

)( ba

is perpendicular to ba

&

Example (Contd)

)ˆˆ3ˆ2()ˆ2ˆˆ5( kjikji

017.29.15.5

)ˆ17ˆ9ˆ5).(ˆ2ˆˆ5(

kjikji

ExampleFind A x B and B x A, if

134

112

kji

-2i-6j+10K

)34(B and ),22( kjikjiA

B x A = 2i+6j-10k, which implies B x A = - (A x B)

Finding a unit vector perpendicular to a plane.

Find a unit vector perpendicular to the plane containing two vectors b&a

Applications

A vector perpendicular to a and b is

Corresponding unit vector

ba

|ba|

ba

kjiba ˆ17ˆ9ˆ5

Cross product

Magnitude 3952898125|| ba

Unit vector is )ˆ17ˆ9ˆ5(395

1kji

ExampleDetermine a unit vector perpendicular

to the plane of

and

)ˆ2ˆˆ5( kjia

)ˆˆ3ˆ2( kjib

Find the area of the triangle with vertices A(1,1,2) B(-1,3,2) and C(4,1,5).Find two sides of the triangle, the vectors AB and AC.

and

Now find the cross product of the two vectors:

The area is given as follows:

Find the area of the triangle with vertices A(1,1,1) B(2,3,4) and C(3,0,-1).

sq unitsArea

kjiPPPP

102

3

2

90

583221

Scalar Triple Product: Example

not.or plane same in the lie

(0,-9,18)c and (2,1,-4)b (1,4,7);a vectors threeif Determine

In here, scalar triple product can be utilized to determine whether the given points lie in same plane or not

plane. same in the lie they implying ,

0

81 9 0

4 1 2

7 4 1

)(

321

321

321

Hence

ccc

bbb

aaa

cba

The 3-dimensional Co-ordinate System

The x-y plane is horizontal in diagram and shaded green. It can also be described using the equation z = 0, since all points on that plane will have 0 for their z-value.

Normally the 'right-hand orientation’ is used for the 3 axes, with the positive x-axis pointing in the direction of the first finger of our right hand, the positive y-axis pointing in the direction of our second finger and the positive z-axis pointing up in the direction of our thumb.

Example The vector OP has initial point at the origin O (0, 0, 0) and terminal point at P (2, 3, 5). We can draw the vector OP as follows:

For the vector OP above, the magnitude of the vector is given by:

| OP | = √(22 + 32 + 52) = 6.16 units

Distance Between 2 Points in 3 Dimensions

If we have point A (x1, y1, z1) and another point B (x2, y2, z2) then the distance AB between them is given by the formula:

Distance AB =sqrt[(x2-x1)2+(y2-y1)2+(z2-z1)2]

Find the distance between the points P (2, 3, 5) and Q (4, -2, 3).

Answer: 5.74 Units

ExampleFind the lengths of the sides of triangle with vertices (0, 0, 0), (5, 4, 1) and (4, -2, 3). Then determine if the triangle is a right triangle, an isosceles triangle or neither.

Solution: First find the length of each side of the triangle by finding thedistance between each pair of vertices.

(0, 0, 0) and (5, 4, 1)

42

11625

010405 222

d

d

d

(0, 0, 0) and (4, -2, 3)

29

9416

030204 222

d

d

d

(5, 4, 1) and (4, -2, 3)

41

4361

134254 222

d

d

d

These are the lengths of the sides of the triangle. Since none of them are equal we know that it is not an isosceles triangle and since Implying it is not a right triangle. Thus it is neither.

222412942

Equation of a Sphere

If we square both sides of this equation we get:

The standard equation of a sphere is

where r is the radius and is the center.

2222r ooo zzyyxx

ooo zyx ,,

A sphere is the collection of all points equal distance from a center point.

To come up with the equation of a sphere, keep in mind that the distance

from any point (x, y, z) on the sphere to the center of the sphere,

is the constant r which is the radius of the sphere.

Using the two points (x, y, z), and r, the radius in the distance

formula, we get:

ooo zyx ,,

222r ooo zzyyxx

ooo zyx ,,

ExercisesFind the equation of the sphere with radius, r = 5 and center, (2, -3, 1).

By Just plugging above values into the standard equation of a sphere we will get:

25)1()3()2( 222 zyx

Find the equation of the sphere with endpoints of a diameter (4, 3, 1) and (-2, 5, 7).

Using the midpoint formula we can find the center and using the distance formula we can find the radius.

4,4,1

271

,2

53,

224

Center

19

919

414314Radius 222

Thus the equation is: 19)4()4()1( 222 zyx

Exercises (Contd)

Find the center and radius of the sphere, . 07864222 zyxzyx

To find the center and the radius we simply need to write the equation of the sphere in standard form, . Then we can easily identify the center, and the radius, r. To do this we will need to complete the square on each variable.

ooo zyx ,,

2222r ooo zzyyxx

36432

169471689644

7864

07864

222

222

222

222

zyx

zzyyxx

zzyyxx

zyxzyx

Thus the center is (2, -3, -4) and the radius is 6.

Adding 3-dimensional Vectors Earlier we saw how to add 2-dimensional vectors. We

now extend the idea for 3-dimensional vectors.

We simply add the i components together, then the j components and finally, the k components.

If A = 2i + 5j − 4k and B = −2i − 3j − 5k, then

determine A+B.

Answer: 2 j − 9 k

Angle Between 3-Dimensional Vectors

Earlier, we saw how to find the angle between 2-dimensional vectors. We use the same formula for 3-dimensional vectors, i.e.

ba

ba

1cos

Find the angle between the vectors a = 4i + 0j + 7k and b = -2i + j + 3k.

a• b = (4 i + 0 j + 7 j) • (-2 i + j + 3 k )

a• b = (4 × -2) + (0 × 1) + (7 × 3)

a• b = 13

Angle Between 3-Dimensional Vectors (contd)

166.3031(-2) 704|| 222222 ba

And now for denominator i.e. magnitudes

So, θ = arccos(13 ÷ 30.166)

Therefore the angle between the vectors a and b is θ = 64.47°

Find the angle between the vectors a = 3i + 4j − 7k and b = -2i + j + 3k.

Answer: θ = 135.6°

ExampleWe have a cube ABCO PQRS which has a string along the cube's diagonal B to S and another along the other diagonal C to P. What is the angle between the 2 strings?

θ

Example (contd) We will assume that we have a unit cube i.e. each side has length 1 unit, with 0 being the origin.

The unit vectors i, j, and k act in the x-, y-, and z-directions respectively. So in our diagram, since we have a unit cube

OA = I, OC = j and OS = k

From the diagram, we see that to move from B to S, we need to go -1 unit in the x direction, -1 unit in the y-direction and up 1 unit in the z-direction. Since we have a unit cube, we can write:

BS = −i − j + k

Similarly, to move from C to PCP = i − j + k

Example (contd)The dot (scalar) product for the vectors BS and CP is:BS • CP = |BS| |CP| cos θ

BS • CP = (−i − j + k) • (i − j + k) = 1

|BS| |CP| = 3

So , θ = arccos (1/3) θ = 70.5°

Therefore, the angle between the strings is 70.5°